In the very fast rotational dynamics of the active particle, the self-propulsion adds up to uncorrelated thermal fluctuations. Thus, following[1]} one can find the expression of mean exit time as,
\(\tau _{ex} = \frac{2R^2}{n(2D_0+l_\theta v_0)}\left[\ln \left(\frac{2\pi R}{\delta \phi }\right)+\frac{n}{8}-\ln \frac{2}{n}\right]. \) 
